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Why is Decimal ('0')> 9999.0 True in Python?

This has something to do with my question Why ``> 0 True in Python?

In Python 2.6.4:

>> Decimal('0') > 9999.0
True

From the answer to my initial question, I understand that when comparing objects of different types in Python 2.x, the types are ordered by their name. But in this case:

>> type(Decimal('0')).__name__ > type(9999.0).__name__
False

Why Decimal('0') > 9999.0 == Truethen?

UPDATE: I usually work on Ubuntu (Linux 2.6.31-20-generiC # 57-Ubuntu SMP Mon Feb 8 09:05:19 UTC 2010 i686 GNU / Linux, Python 2.6.4 (r264: 75706, December 7, 2009, 18 : 45: 15) [GCC 4.4.1] on linux2). On Windows (WinXP Professional SP3, Python 2.6.4 (r264: 75706, November 3, 2009, 13:23:17) [MSC v.1500 32 bit (Intel)] on win32) my original expression works differently:

>> Decimal('0') > 9999.0
False

I am even more puzzled. % - (

+5
2

, long, int Decimal. " , " . _convert_other() decimal.py

, .

, . http://bugs.python.org/issue2531.

, :

  • Decimal.__gt__.
  • Decimal.__gt__ Decimal._convert_other, float .
  • Decimal._convert_other . Decimal._convert_other long, int Decimal . , . TypeException. NotImplemented, , , Employee.
  • . .
  • CPython Objects/object.c/default_3way_compare.
  • Python 3 . Python 2 id().
  • Windows (). .
  • , .

?

+12
def __gt__(self, other, context=None):
    other = _convert_other(other)
    if other is NotImplemented:
        return other
    ans = self._compare_check_nans(other, context)
    if ans:
        return False
    return self._cmp(other) > 0


def _convert_other(other, raiseit=False):
    """Convert other to Decimal.

    Verifies that it ok to use in an implicit construction.
    """
    if isinstance(other, Decimal):
        return other
    if isinstance(other, (int, long)):
        return Decimal(other)
    if raiseit:
        raise TypeError("Unable to convert %s to Decimal" % other)
    return NotImplemented
+1

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